Find the first derivative for the following function.?

Question

1)f(x)=x -1/2x+5

2)f(x) = (3x^2 -1)(2x^3 - 5)

3)f(x)= 3 square root1-3x

Answer 1

some of your statements are kind of poorly stated, so I have to make guesses at them.

1)f(x) = x - (1/2)x+5

f' = 1 - 1/2 = 1/2

2)f(x) = (3x² -1)(2x³ - 5)

f(x) = 6x⁵ -  2x³ + 5 - 15x²

f' = 30x⁴ -  6x² - 20x

3)f(x) = 3*√1 - 3x

f(x) = 3 - 3x

f' = 3

.

Answer 2

1) 1-1/2

     1/2

2) 6x^5-2x^3-15x^2+5 (After you FOIL it.)

     30x^4-6x^2-30X

3)3*√1 - 3x

   3 - 3x

    3

Answer 3

Would u attempt ur own HW , it's not that difficult

Answer 4

For the first function, I do not think that you meant

(1) f(x)=x-0.5x+5 which = 0.5x+5

and the first derivative would have been 0.5.

I  think  u really meant to write f(x)=x-1/(2x)+5

then the first derivative is 1+1/(2x^2)

For the second function,

(2) f(x) = (3x^2 -1)(2x^3 - 5) which in expanded form is

6x^5-2x^3-15x^2+5

the first derivative is 30x^4-6x^2-30x

For the third function,

3)f(x)= 3 square root1-3x

I think u really meant

(3) f(x)=3*(1-3x)^0.5

the first derivative is -4.5*(1-3x)^-0.5

But if u meant to write f(x)=3*(1)^0.5-3x

which is equivalent to f(x)=3-3x

then the first derivative is -3.

Hope i helped ya with your problem. Just one tip if ya want to get accurate answers, pls use brackets to make your ? clear.

Gd luck.

Answer 5

1.    f(x) =  (x--1)/(2x+5)

       dy/dx =  [(2x+5) .d/dx(x--1) -- (x--1) d/dx (2x+5)]/(2x+5)^2

                 =  {(2x+5).1 -- (x-1).2}/(2x+5)^2

                  =   (4x+7)/ (2x+5)^2  

3.   f(x)   =  3 Sqroot (1--3x)

      dy/dx=   (1/2)(1--3x)^1/2-1muliplied by --3

               =  (--3)/{2 Sqroot(1--3x)}

2.    f(x) =  (3x^2--1)(2x^3--5)

      dy/dx=   (3x^2--1) .6x^2 + (2x^3--5) . 6x

   On simplification, we get

               = (30x^4 --6x^2 --30x)